Is 1 + 1/2 + ... + 1/n in the denominator?
If it is in the numerator, then this is divergent, since it is greater than 1/n, which is divergent.
(EDIT: nevermind this. I thought it was a sum!)
If 1 + 1/2 + ... + 1/n is in the denominator, then the denominator itself is increasing (at each step the sum gets bigger, and it gets multiplier by a greater number). This implies that the sequence is decreasing. It is bounded below since all its terms are positive. And it will converge to 0, since it is smaller than 1/n, which converges to 0.